The frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.
<h3>What is a frequency?</h3>
The number of waves that travel through a particular point in a given length of time is described by frequency. So, if a wave takes half a second to pass, the frequency is 2 per second.
Given that the energy of the photon is 4.56 x 10⁻¹⁹ J. Therefore, the frequency of the photon can be written as,
![\rm \gamma = \dfrac{E}{h} = \dfrac{4.56x10^{-19} J}{6.626 \times 10^{-34}\ Jsec^{-1}}\\\\\\\gamma = 6.88 \times 10^{14}\ s^{-1}](https://tex.z-dn.net/?f=%5Crm%20%5Cgamma%20%3D%20%5Cdfrac%7BE%7D%7Bh%7D%20%3D%20%5Cdfrac%7B4.56x10%5E%7B-19%7D%20J%7D%7B6.626%20%5Ctimes%2010%5E%7B-34%7D%5C%20Jsec%5E%7B-1%7D%7D%5C%5C%5C%5C%5C%5C%5Cgamma%20%20%3D%206.88%20%5Ctimes%2010%5E%7B14%7D%5C%20s%5E%7B-1%7D)
Hence, the frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.
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Answer:
A. False, frequency can increase or decrease wavelength.
For example: a high frequency would mean there are shorter wavelengths that occur in a period. Meanwhile, a low frequency would indicate that the wavelengths are longer and in longer periods.
Answer:
Options 1 and 5 are correct
Explanation:
Magnetic field lines can never cross, the field is unique at any point in space. Magnetic field lines are continuous, forming closed loops without beginning or end. They go from the north pole to the south pole.
Magnetic field lines form closed loops but do not intersect.
Electric field lines originate at the positive charges and terminate at the negative charges. They move in a straight line and are parallel. Electric field lines neither form closed loops nor intersect.
Since, magnetic field lines form closed loops and move from North to South pole, they come out of north poles outside the magnet and into north poles inside the magnet, they also go into south poles outside the magnet and out of south poles inside the magnet.
It often requires STANDARD reaction time
Answer:
C = 1.01
Explanation:
Given that,
Mass, m = 75 kg
The terminal velocity of the mass, ![v_t=60\ m/s](https://tex.z-dn.net/?f=v_t%3D60%5C%20m%2Fs)
Area of cross section, ![A=0.33\ m^2](https://tex.z-dn.net/?f=A%3D0.33%5C%20m%5E2)
We need to find the drag coefficient. At terminal velocity, the weight is balanced by the drag on the object. So,
R = W
or
![\dfrac{1}{2}\rho CAv_t^2=mg](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7D%5Crho%20CAv_t%5E2%3Dmg)
Where
is the density of air = 1.225 kg/m³
C is drag coefficient
So,
![C=\dfrac{2mg}{\rho Av_t^2}\\\\C=\dfrac{2\times 75\times 9.8}{1.225\times 0.33\times (60)^2}\\\\C=1.01](https://tex.z-dn.net/?f=C%3D%5Cdfrac%7B2mg%7D%7B%5Crho%20Av_t%5E2%7D%5C%5C%5C%5CC%3D%5Cdfrac%7B2%5Ctimes%2075%5Ctimes%209.8%7D%7B1.225%5Ctimes%200.33%5Ctimes%20%2860%29%5E2%7D%5C%5C%5C%5CC%3D1.01)
So, the drag coefficient is 1.01.